# A compound is 54.53% C, 9.15% H, and 36.32% O by mass. What is its empirical formula?

Feb 26, 2016

${C}_{2} {H}_{4} O$

#### Explanation:

The ratio of no of atoms of C,H&O=$\left(\frac{54.53}{12}\right) : \left(\frac{9.15}{1}\right) : \left(\frac{36.32}{16}\right)$

=$4.54 : 9.15 : 2.27$

=$\frac{4.54}{2.27} : \frac{9.15}{2.27} : \frac{2.27}{2.27}$

=$2 : 4 : 1$

Empirical formula ${C}_{2} {H}_{4} O$